# How come, for any n > 2, the nth triangular number + the nth square

number cannot be prime? I have checked this for n from 3 to 53,509, the latter being the limit for unsigned int. I believe this is true, and I thereby claim that this is a true statement. However, I don't see any obvious explanation for it.

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Mentallic
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The nth triangle number Tn is

$$T_n = \frac{n(n+1)}{2}$$

and the nth square number Sn is

$$S_n = n^2$$

And so
$$T_n+S_n = \frac{n(n+1)}{2}+n^2=\frac{n(3n+1)}{2}$$

Can you now show why this expression must be composite (not prime) for all n?

Can you now show why this expression must be composite (not prime) for all n?
Well, it wouldn't be true for all n. Only for n > 2, assuming my statement is correct.

When n is odd and more than 2, 1 / 2 * (3n + 1) is a whole number bigger than 1, and so the result is composite. When n is even and more than 2, n * (3n + 1) is even, so (n / 2) * (3n + 1) is even. Therefore the sum is composite for all n > 2 is correct. Many thanks for the help.

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Mentallic
Homework Helper
Well, it wouldn't be true for all n. Only for n > 2, assuming my statement is correct.
Well, ignoring the fact that you included the criteria that n>2 in your proof below, $n\leq 2$ would also spit out composite numbers, right? Your proof only considers that n is even or odd which means that all integers $n\leq 2$ would also be involved. The only reason to restrict yourself to n>2 is such that we have a meaningful square and triangle number.

When n is odd and more than 2, 1 / 2 * (3n + 1) is a whole number bigger than 1, and so the result is composite.
Adding to the end of that: because we then have a product of two integers, mainly n and $\frac{3n+1}{2}$.

When n is even and more than 2, n * (3n + 1) is even, so (n / 2) * (3n + 1) is even. Therefore the sum is composite for all n > 2 is correct. Many thanks for the help.
This is incorrect. If $n(3n+1)$ is even, then $\frac{n}{2}(3n+1)$ isn't necessarily even, but rather an integer. But most importantly, you haven't proven that the expression is a product of two integers and hence composite.

Also, while it's not absolutely necessary, when you consider n to be even, you could let n=2k for some integer k and substitute that into your expression, then show that the result is composite, and similarly for n odd, let n=2k+1.

Many thanks.

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Well, ignoring the fact that you included the criteria that n>2 in your proof below, $n\leq 2$ would also spit out composite numbers, right? Your proof only considers that n is even or odd which means that all integers $n\leq 2$ would also be involved. The only reason to restrict yourself to n>2 is such that we have a meaningful square and triangle number.
You do get a prime number for n=1 or n=2, and the proof also uses the fact that n>2, when it says
When n is odd and more than 2, 1 / 2 * (3n + 1) is a whole number bigger than 1,
so I don't see what the problem is here.

Mentallic
Homework Helper
Haha yeah I thought about it while out today and realized the criteria n>2 is necessary, which goldust even incorporated into his proof! Sorry about that goldust.

Many thanks for the help! The proof is a bit trickier than I initially thought. When n is even and more than 2, n / 2 is an integer more than 1, and 3n + 1 is also an integer more than 1, so n / 2 * (3n + 1) ends up being divisible by both n / 2 and 3n + 1. Cheers!

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